Factor polynomial $$ 6x^{2}-17x+12 $$

$$ 6x^{2}-17x+12 = ( 2x-3 )( 3x-4 ) $$

Step 1: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 6 , b = -17 ~ \text{ and } ~ c = 12 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 6 }$ by the constant term $\color{blue}{c = 12} $.

$$ a \cdot c = 72 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 72 $ and add to $ b = -17 $.

Step 4: All pairs of numbers with a product of $ 72 $ are:

PRODUCT = 72
1     72	-1     -72
2     36	-2     -36
3     24	-3     -24
4     18	-4     -18
6     12	-6     -12
8     9	-8     -9

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -17 }$

PRODUCT = 72 and SUM = -17
1     72	-1     -72
2     36	-2     -36
3     24	-3     -24
4     18	-4     -18
6     12	-6     -12
8     9	-8     -9

Step 6: Replace middle term $ -17 x $ with $ -8x-9x $:

$$ 6x^{2}-17x+12 = 6x^{2}-8x-9x+12 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 6x^{2}-8x-9x+12 = 2x\left(3x-4\right) -3\left(3x-4\right) = \left(2x-3\right) \left(3x-4\right) $$

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